3.8.100 \(\int \frac {(a+b x)^2}{x^3 (c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=66 \[ -\frac {a^2}{5 c x^4 \sqrt {c x^2}}-\frac {a b}{2 c x^3 \sqrt {c x^2}}-\frac {b^2}{3 c x^2 \sqrt {c x^2}} \]

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Rubi [A]  time = 0.01, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {15, 43} \begin {gather*} -\frac {a^2}{5 c x^4 \sqrt {c x^2}}-\frac {a b}{2 c x^3 \sqrt {c x^2}}-\frac {b^2}{3 c x^2 \sqrt {c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^2/(x^3*(c*x^2)^(3/2)),x]

[Out]

-a^2/(5*c*x^4*Sqrt[c*x^2]) - (a*b)/(2*c*x^3*Sqrt[c*x^2]) - b^2/(3*c*x^2*Sqrt[c*x^2])

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[(a^IntPart[m]*(a*x^n)^FracPart[m])/x^(n*FracPart[m]), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin {align*} \int \frac {(a+b x)^2}{x^3 \left (c x^2\right )^{3/2}} \, dx &=\frac {x \int \frac {(a+b x)^2}{x^6} \, dx}{c \sqrt {c x^2}}\\ &=\frac {x \int \left (\frac {a^2}{x^6}+\frac {2 a b}{x^5}+\frac {b^2}{x^4}\right ) \, dx}{c \sqrt {c x^2}}\\ &=-\frac {a^2}{5 c x^4 \sqrt {c x^2}}-\frac {a b}{2 c x^3 \sqrt {c x^2}}-\frac {b^2}{3 c x^2 \sqrt {c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 33, normalized size = 0.50 \begin {gather*} \frac {c \left (-6 a^2-15 a b x-10 b^2 x^2\right )}{30 \left (c x^2\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^2/(x^3*(c*x^2)^(3/2)),x]

[Out]

(c*(-6*a^2 - 15*a*b*x - 10*b^2*x^2))/(30*(c*x^2)^(5/2))

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IntegrateAlgebraic [A]  time = 0.03, size = 35, normalized size = 0.53 \begin {gather*} \frac {-6 a^2-15 a b x-10 b^2 x^2}{30 x^2 \left (c x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + b*x)^2/(x^3*(c*x^2)^(3/2)),x]

[Out]

(-6*a^2 - 15*a*b*x - 10*b^2*x^2)/(30*x^2*(c*x^2)^(3/2))

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fricas [A]  time = 1.28, size = 34, normalized size = 0.52 \begin {gather*} -\frac {{\left (10 \, b^{2} x^{2} + 15 \, a b x + 6 \, a^{2}\right )} \sqrt {c x^{2}}}{30 \, c^{2} x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2/x^3/(c*x^2)^(3/2),x, algorithm="fricas")

[Out]

-1/30*(10*b^2*x^2 + 15*a*b*x + 6*a^2)*sqrt(c*x^2)/(c^2*x^6)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (b x + a\right )}^{2}}{\left (c x^{2}\right )^{\frac {3}{2}} x^{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2/x^3/(c*x^2)^(3/2),x, algorithm="giac")

[Out]

integrate((b*x + a)^2/((c*x^2)^(3/2)*x^3), x)

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maple [A]  time = 0.00, size = 32, normalized size = 0.48 \begin {gather*} -\frac {10 b^{2} x^{2}+15 a b x +6 a^{2}}{30 \left (c \,x^{2}\right )^{\frac {3}{2}} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^2/x^3/(c*x^2)^(3/2),x)

[Out]

-1/30*(10*b^2*x^2+15*a*b*x+6*a^2)/x^2/(c*x^2)^(3/2)

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maxima [A]  time = 1.36, size = 33, normalized size = 0.50 \begin {gather*} -\frac {b^{2}}{3 \, c^{\frac {3}{2}} x^{3}} - \frac {a b}{2 \, c^{\frac {3}{2}} x^{4}} - \frac {a^{2}}{5 \, c^{\frac {3}{2}} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2/x^3/(c*x^2)^(3/2),x, algorithm="maxima")

[Out]

-1/3*b^2/(c^(3/2)*x^3) - 1/2*a*b/(c^(3/2)*x^4) - 1/5*a^2/(c^(3/2)*x^5)

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mupad [B]  time = 0.20, size = 42, normalized size = 0.64 \begin {gather*} -\frac {6\,a^2\,\sqrt {x^2}+10\,b^2\,x^2\,\sqrt {x^2}+15\,a\,b\,x\,\sqrt {x^2}}{30\,c^{3/2}\,x^6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^2/(x^3*(c*x^2)^(3/2)),x)

[Out]

-(6*a^2*(x^2)^(1/2) + 10*b^2*x^2*(x^2)^(1/2) + 15*a*b*x*(x^2)^(1/2))/(30*c^(3/2)*x^6)

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sympy [A]  time = 0.98, size = 56, normalized size = 0.85 \begin {gather*} - \frac {a^{2}}{5 c^{\frac {3}{2}} x^{2} \left (x^{2}\right )^{\frac {3}{2}}} - \frac {a b}{2 c^{\frac {3}{2}} x \left (x^{2}\right )^{\frac {3}{2}}} - \frac {b^{2}}{3 c^{\frac {3}{2}} \left (x^{2}\right )^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**2/x**3/(c*x**2)**(3/2),x)

[Out]

-a**2/(5*c**(3/2)*x**2*(x**2)**(3/2)) - a*b/(2*c**(3/2)*x*(x**2)**(3/2)) - b**2/(3*c**(3/2)*(x**2)**(3/2))

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